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Question: For the principal value, evaluate the following \[{{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)...

For the principal value, evaluate the following
sin⁡−1(32)−2sec⁡−1(2tan⁡π6){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( 2\tan \dfrac{\pi }{6} \right)

Explanation

Solution

Hint:First of all, substitute the value of tan⁡π6\tan \dfrac{\pi }{6} from the trigonometric table. Now find the angles at which sin⁡θ=32\sin \theta =\dfrac{\sqrt{3}}{2} and secθ=23sec\theta =\dfrac{2}{\sqrt{3}} or the value of sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) and sec−1(23)se{{c}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) and substitute these in the given expression to get the required answer.

Complete step-by-step answer:
In this question, we have to find the principal value of the expression sin⁡−1(32)−2sec⁡−1(2tan⁡π6){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( 2\tan \dfrac{\pi }{6} \right).
First of all, let us consider the expression given in the question,
E=sin⁡−1(32)−2sec⁡−1(2tan⁡π6)....(i)E={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( 2\tan \dfrac{\pi }{6} \right)....\left( i \right)
Now, let us draw the table for trigonometric ratios of general angles.

From the above table, we get, tan⁡π6=13\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}. So, by substituting the value of tan⁡π6\tan \dfrac{\pi }{6} in equation (i), we get,
E=sin⁡−1(32)−2sec⁡−1(2.13)E={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( 2.\dfrac{1}{\sqrt{3}} \right)
E=sin⁡−1(32)−2sec⁡−1(23).....(ii)E={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right).....\left( ii \right)
Now we know that the range of principal value of sin⁡−1(x){{\sin }^{-1}}\left( x \right) lies between [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
From the table of general trigonometric ratios, we get,
sin⁡(π3)=32\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}
By taking sin⁡−1{{\sin }^{-1}} on both the sides, we get,
sin⁡−1sin⁡(π3)=sin⁡−1(32){{\sin }^{-1}}\sin \left( \dfrac{\pi }{3} \right)={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)
We know that for −π2≤x≤π2,sin⁡−1sin⁡(x)=x\dfrac{-\pi }{2}\le x\le \dfrac{\pi }{2},{{\sin }^{-1}}\sin \left( x \right)=x. So, we get,
π3=sin⁡−1(32)...(iii)\dfrac{\pi }{3}={{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)...\left( iii \right)
Now, we also know that the range of principal value of sec−1xse{{c}^{-1}}x lies between \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}
From the table of trigonometric ratios, we get,
sec(π6)=23sec\left( \dfrac{\pi }{6} \right)=\dfrac{2}{\sqrt{3}}
By taking sec−1se{{c}^{-1}} on both the sides, we get,
sec−1(sec(π6))=sec−1(23)se{{c}^{-1}}\left( sec\left( \dfrac{\pi }{6} \right) \right)=se{{c}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)
We know that for \left[ 0\le x\le \pi \right]-\left\\{ \dfrac{\pi }{2} \right\\},se{{c}^{-1}}sec\left( x \right)=x. So, we get,
π6=sec−1(23)....(iv)\dfrac{\pi }{6}=se{{c}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)....\left( iv \right)
So, by substituting the value of sec−1(23)se{{c}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right) from equation (iv) and sin⁡−1(32){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right) from equation (iii) in equation (ii), we get,
E=π3−2(π6)E=\dfrac{\pi }{3}-2\left( \dfrac{\pi }{6} \right)
E=π3−π3E=\dfrac{\pi }{3}-\dfrac{\pi }{3}
E=0E=0
Hence, we get the value of sin⁡−1(32)−2sec⁡−1(2tan⁡π6){{\sin }^{-1}}\left( \dfrac{\sqrt{3}}{2} \right)-2{{\sec }^{-1}}\left( 2\tan \dfrac{\pi }{6} \right) as 0.

Note: In these types of questions, first of all, students must remember the trigonometric ratios at general angles like 0o,30o,45o,60o,90o,{{0}^{o}},{{30}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}}, etc. Also, students must note that sin⁡−1sin⁡(x)=x{{\sin }^{-1}}\sin \left( x \right)=x and sec−1sec⁡(x)=xse{{c}^{-1}}\sec \left( x \right)=x are the only value for their respective ranges and not for other values of x. Students should not confuse between sin⁡−1sin⁡x{{\sin }^{-1}}\sin x and sin⁡sin⁡−1x\sin {{\sin }^{-1}}x and simplify for other trigonometric ratios as well.